Help physics?
2016-07-31 2:09 am
A railway Wagon of mass 2.50 tonnes moving along a horizontal track at 2 m/s runs into a stationary engine and is coupled to it. After the collision, the engine and wagon move off to a slow 0.300 m/s. What is the mass of the engine alone?
回答 (4)
2016-07-31 7:16 am
prior hooking
I = mw*Vw = 2.5*2 = 5.0
after hooking I is maintained, then ;
V = I/(mw+me)
0.3*2.5+0.3*me = 5.0
0.3me = 5-0.75
me = 4.25/0.3 = 42.5/3 = 14.1(6) tons
I = mw*Vw = 2.5*2 = 5.0
after hooking I is maintained, then ;
V = I/(mw+me)
0.3*2.5+0.3*me = 5.0
0.3me = 5-0.75
me = 4.25/0.3 = 42.5/3 = 14.1(6) tons
2016-07-31 2:25 am
One tonne is 1000 kg
Mass = 2500 kg
Initial momentum = 2500 * 2 = 5000
Total mass = 2500 + m
Final momentum = (2500 + m) * 0.3 = 750 + 0.3 * m
750 + 0.3 * m = 5000
0.3 * m = 4250
m = 4250 ÷ 0.3 = 14,166⅔ kg
Mass = 2500 kg
Initial momentum = 2500 * 2 = 5000
Total mass = 2500 + m
Final momentum = (2500 + m) * 0.3 = 750 + 0.3 * m
750 + 0.3 * m = 5000
0.3 * m = 4250
m = 4250 ÷ 0.3 = 14,166⅔ kg
2016-07-31 2:20 am
...
this is an inelastic collision
M1V1 + M2V2 = (M1_M2) V3
we may o ahead and use mass in tonnes
2.50 (2) + 0 = (2.5+X) (0.300)
5.00 = 0.75 + 0.300 X
0.300 X = 4.25
X = 14.2 tonnes
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this is an inelastic collision
M1V1 + M2V2 = (M1_M2) V3
we may o ahead and use mass in tonnes
2.50 (2) + 0 = (2.5+X) (0.300)
5.00 = 0.75 + 0.300 X
0.300 X = 4.25
X = 14.2 tonnes
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
You may have to wait an hour to award BA.
2016-07-31 2:19 am
mass of railway car {US term for it} = 2.50 tonnes
initial speed of railway car = 2 m/s
initial momentum = 2(2.5) = 5 units {m-tonne/s}
x = mass of engine {locomotive engine} in tonnes
after collision mass = (2.50 + x)
after collision speed = 0.300 m/s
after collision momentum = 0.300(2.50 + x) = 0.750 + 0.300x
Conservation of Momentum
{as both before & after momentums are in same direction}
5 = 0.750 + 0.300x
x = 4.25/0.300 = 14.2 tonnes ANS
initial speed of railway car = 2 m/s
initial momentum = 2(2.5) = 5 units {m-tonne/s}
x = mass of engine {locomotive engine} in tonnes
after collision mass = (2.50 + x)
after collision speed = 0.300 m/s
after collision momentum = 0.300(2.50 + x) = 0.750 + 0.300x
Conservation of Momentum
{as both before & after momentums are in same direction}
5 = 0.750 + 0.300x
x = 4.25/0.300 = 14.2 tonnes ANS
收錄日期: 2021-04-21 19:29:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160730180932AAs7eY8